Manifold orientable iff nth exterior power of the cotangent bundle is trivial

Question

Let $M$ be a manifold with dimension $n$. Then prove $M$ is orientable if and only if $\Lambda^nT^*M$ is trivial.

For both directions I used the argument of existence of global frame implied by the hypothesis. However it does not seem right.

Answer 1

Suppose $\bigwedge^n T^\ast M$ is trivial (vector bundle). That means that there is a $n$-form $\omega$ such that $\omega_p \neq 0$ for each $p\in M$. For each $p \in M$ take a chart $x:U \to \mathbb R^n$, with $p\in U$, such that $$\omega = a dx_1 \wedge \ldots \wedge dx_n$$ with $a(q)>0$ for each $q\in U$. This family of charts we chose give us a oriented atlas.

On the other hand, suppose we have a oriented atlas. For each chart $x:U \to \mathbb R^n$ on this atlas we obtain de $n$-form $dx_1 \wedge \ldots \wedge dx_n$ defined on $U$. Using partition of unity you can glue this $n$-forms in order to produce a $n$-form $\omega$ such that $\omega_p\neq 0$ for each $p\in M$. That means that $\bigwedge^n T^\ast M$ is trivial.